I was trying to solve some exam problems when I encountered a really peculiar exercise, that I couldn't figure out, unfortunately: Find the function $f(x) s.t. f(x) \to 0$ as $|x|\to \infty$, which satisfies the equation (integrals are all from $-\infty,\infty$).
$$\iiint f(y)g_1(z)g_1(t)\delta(x-y-z-t)dydzdt = g_2(x)$$
where $$g_1(x) = \frac{e^{-\frac{(x-m)^2}{2\alpha^2} } }{\sqrt{2\pi\alpha^2}},\quad g_2(x) = \frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}.$$
The solution says that the first triple integral represents the convolution between $f(x) \ast G(x-y)$, where $G(x-y) = g_1 \ast g_1$, namely the convolution of $g_1$ with itself. After having seen this I was able to fully solve the exercise. The problem still remains: how can a human recognize that the first triple integral is a convolution of a convolution. Is it possible to show it in some way, maybe using the Dirac delta properties? I can't come up with a way of putting the first triple integral in the usual form of a convolution ( $\int f(y) G(x-y)dy$, which the solution gives me). I think the delta is there to somehow get rid of two of the integrals (in $dz$ and $dt$). I was thinking of a clever change of variable but couldn't come up with it. Can someone help me with this? Every help is well appreciated ;)